Solving Sin/sec/tan/csc: -sqrt6/6

  • Context: Undergrad 
  • Thread starter Thread starter sfeld
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around a trigonometric expression involving angles in both degrees and radians, specifically evaluating the expression sin(240°) × sec(-45°) - tan(-π/6) × csc(-315°). Participants express confusion over the mixing of angle measures and the implications for solving the problem.

Discussion Character

  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant asks how to arrive at the answer of -√6/6 for the given trigonometric expression.
  • Another participant suggests converting all angles to radians and using reference triangles to find the values of each term, indicating that the problem is basic if approached this way.
  • Some participants express frustration over the mixing of degrees and radians in the problem, questioning the clarity and appropriateness of the notation used.
  • There are comments on the misunderstanding of radian and degree functions, with some participants arguing that they yield different values for the same input.
  • One participant humorously suggests that the author of the problem may be intentionally trying to confuse others.

Areas of Agreement / Disagreement

Participants generally agree that the mixing of degrees and radians is problematic, but there is no consensus on the best approach to solve the problem or on the intent behind the problem's formulation.

Contextual Notes

Participants note the potential confusion arising from the use of both degrees and radians without clear indication, as well as the implications for understanding trigonometric functions.

sfeld
Messages
12
Reaction score
0
How do they figure this answer out?

sin 240 x sec(-45) - tan (-pie/6) csc(-315)

Basiclaly it says

sin(240 degrees) times sec(-145 degrees) - tan(-pie/positive 6) csc(-315 degrees)

says the answer is -square root 6/6

sorry I don't know how to do the symbol stuff on this forum :( please forgive me.
 
Physics news on Phys.org
It is not the policy of this site do homework for you. You'll have to figure it out for yourself. I'll give you a hint, however. Change all of the angles into radians, and using reference triangles, just figure out the values of each term directly. (You'll know the ratios because all the triangles are either 30º-60º-90º or 45º-45º-90º triangles). This is the most basic way to do it.

On a side note, [itex]\pi[/itex] is spelled "pi," not "pie." A pie is something that you eat. Pi is the Greek letter.
 
Last edited:
Am I the only one that thinks that a trig problem in which angles are given as both degrees (45) and radians (-pi/6) is really horrible? Especially if there was not mention as to which was which. If I saw "-pi/6" here, why wouldn't I assume that was radians and that "sec -45" means the secant of -45 radians?
 
HallsofIvy said:
Am I the only one that thinks that a trig problem in which angles are given as both degrees (45) and radians (-pi/6) is really horrible?
It's a blasphemy!
In addition, the author has evidently not understood that "radian trig functions" are different functions than "degree trig functions", since, given the same input yield different values.
 
Last edited:
arildno said:
In addition, the author has evidently not understood that "radian trig functions" are different functions than "degree trig functions", since, given the same input yield different values.

I think the author is just trying to confuse these guys even further. They don't need that.
 
In that case, he's a sadist; that's even worse than ignorance, IMO.
 

Similar threads

Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K